在arrayList中删除（index）不删除该索引处的对象

Question

当执行进入非Reps.remove行时，删除不是删除指定索引处的对象。为什么？

//given array
Character[] arr = new Character[]{'a','a','a','b','c','c','c','d','e','e','e','f'};


Map<Character,Integer> map = new HashMap();
List<Character> nonReps = new ArrayList<>();

for (int i = 0; i < arr.length; i++) {
         if(map.containsKey(arr[i])){
                Integer indexInNonreps = map.get(arr[i]);
                Character characterInNonreps = nonReps.get(indexInNonreps);
                if(arr[i].equals(characterInNonreps))
                    nonReps.remove(indexInNonreps);
         }else{
                nonReps.add(arr[i]);
                map.put(arr[i],nonReps.size()-1);
          }
}

System.out.println(nonReps);

此解决方案打印：[a， b， c， d， e， f]当它应该是b， d， f时

Answer 1

您可以使用Set检查字符是否唯一，如果不是，则从noReps中删除

    public static void main(String[] args) {
        Character[] arr = new Character[]{'a','a','a','b','c','c','c','d','e','e','e','f'};

        Set<Character> seen = new HashSet<>();
        List<Character> nonReps = new ArrayList<>();

        for (int i = 0; i < arr.length; i++) {
            if(seen.contains(arr[i])){
                nonReps.remove(arr[i]);
            }else{
                nonReps.add(arr[i]);
                seen.add(arr[i]);
            }
        }
        System.out.println(nonReps);
    }

，输出

[b, d, f]